Question

The Pear company sells pPhones. The cost to manufacture x pPhones is C(x)=-27x2+65000x+21818 dollars (this includes overhead costs and production costs for each pPhone). If the company sells x pPhones for the maximum price they can fetch, the revenue function will be R(x)=-36x2+353000x dollars.

How many pPhones should the Pear company produce and sell to maximimze profit? (Remember that profit=revenue-cost.)

Answer 1

To maximize the profit, we need to find the value of
`x` that maximizes the difference between the revenue
`R(x)` and the cost
`C(x)`, which gives us the profit function
`P(x) = R(x) - C(x)`.

Given:

`C(x) = -27x^(2) + 65000x + 21818\\R(x) = -36x^(2) + 353000x`

The profit function is:

`P(x) = R(x) - C(x) = (-36x^(2) + 353000x) - (-27x^(2) +65000x + 21818)\\P(x) = -9x^(2) + 288000x - 21818`

To find the value
`x` that maximizes the profit, we need to find the vertex of the parabolic profit function
`P(x)`. The
`x`-coordinate of the vertex is given by:

`x=-(b)/(2a)`

In our case,
`a = -9` and
`b = 288000,` so:

`x=-(288000)/(2(-9)) = 8000`

Now we can find the corresponding
`P(x)` value:

`P(8000)=-9(8000)^(2) + 288000`×
`8000 - 21818`

Calculating this:

`P(8000) = -57600000 + 230400000 - 21818 = 172821182`

Therefore, the maximum profit occurs when the Pear company produces and sells 8000 pPhones, and the maximum profit is $172,821,182.